The quantity of electrolysis products formed depends on:
- Quantity of electric charge passed through the electrolyte
- Quantity of charge required to produce one mole of the substance.
Faraday’s first law of electrolysis: The amount of the substance produced at the electrode is directly proportional to the amount of the electric charge (electric current times the time of current passed) passed through the electrolyte.
- Mathematically, the current in amperes is equal to the charge in coulombs divided by time in seconds.
- The greater the electric current applied and the longer time the current is passed, the greater the quantity of charge passed, which in turn, results in the greater amount of substance discharged at the electrodes.
- The quantity of charge to produce one mole of the substance is the charge carried by the electrons lost or gained during the discharge of 1 mole of ions at the electrodes.
- It depends on the stoichiometry of the half-reaction. That is the mole ratio of the ion and electron involved in the half-reaction.
  - From the half-reactions, we can observe that the number of moles of electrons needed to produce 1 mole of a product at an electrode is equal to the charge number of the ion.
  - The greater the charge number of an ion, the greater the number of electrons involved to discharge the ion. The greater the number of electrons involved, the greater the charge required to discharge 1 mole of the ion.

Faraday’s second law of electrolysis: The amount of substance produced is inversely proportional to the charge on the ion.
- The amount of substance deposited depends on:
  - Current passed
  - Time of current passed
  - Charge on the ion
Sample Question
How much of the charge will be required to deposit 1 mole of aluminium from its ion?

Faraday’s constant:
- The product of Avogadro’s number and the electronic charge is the Faraday constant. The Faraday constant is the charge carried by one mole of electrons in coulombs. It is used to calculate the charge required to deposit 1 mole of the product.
- Faraday constant (F) = 9.65×10⁴ C mol^–1
- 96500 C = 1 Faraday
- 1 mole of aluminium needs a charge of 3 faradays, 1 mole of silver needs 1 Faraday and 1 mole of copper needs 2 faradays to be deposited.
- The charge required to deposit one mole of an element is the number of faradays equal to the number of charges on its ion.
- Sample Question: A 40.0 ampere current flows through molten iron(III) chloride for 10.0 hours (36,500 s). What is the correct mass of iron that is produced during this time? (A_r of iron = 55.8))
- Sample Question: Calculate the volume of chlorine gas produced at rtp in the above electrolysis.
- Sample Question: How long would it take to deposit 0.635 g of copper at the cathode during the electrolysis of copper(II) sulfate solution if you use a current of 0.200 A?
  F= 9.65 x 10⁴ C mol^-1 (or 96500 C mol^-1), A_r of Cu = 63.5.

The map of quantitative electrolysis:
- This can be used to solve the value of:
  - Faraday constant (F)
  - Avogadro’s constant (A)
  - Electronic charge (e)
- Sample Question: A constant electric current of 0.20 A was passed for 34 minutes through a solution of aqueous copper(II) sulfate. The mass of the anode decreased by 0.13 g. Use this data to calculate Avogadro constant.
  Ar value: [Cu] = 63.5, Charge of an electron (e)= 1.6×10^-19C

4.02 - Electrolysis (II)

Learn Step-By-Step Video Explanation For This Topic

Explore our Revision Resources

Mock Exams

Tutor Support

Create an account